Problem: $\overline{AB}$ = $\sqrt{65}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $\sqrt{65}$ $?$ $ \sin( \angle BAC ) = \frac{4\sqrt{65} }{65}, \cos( \angle BAC ) = \frac{7\sqrt{65} }{65}, \tan( \angle BAC ) = \dfrac{4}{7}$
$\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{\sqrt{65}} $ $ \overline{BC}=\sqrt{65} \cdot \sin( \angle BAC ) = \sqrt{65} \cdot \frac{4\sqrt{65} }{65} = 4$